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\title{Algebraic Number Theory, PS 1}
\author{Bryden Cais}
\begin{document}
\maketitle
\begin{enumerate}

    \item Let $L/K$ be an extension of number fields.  It is not true that for $\alpha\in L$, $\tr(\alpha),\n(\alpha)\in \O_K$ implies
    $\alpha\in \O_L.$  As a counterexample, set $K=\Q$ and let $\alpha$ be a root of
    $$f(x)=x^3+\frac{1}{3}x-1$$ and let $L=K(\a)$ Then we have $\tr(\a)=0,\n(\a)=1.$  It is clear that $f$ is irreducible: if it were not,
    it would have a rational root.  But such a root would have numerator and denominator dividing 3, which (after a brief search) is impossible.
    It follows that $f$ is the minimal monic polynomial of $\a$ so that $\a$ is not integral.

    If $\deg(L/K)=2$ then $\tr_{L/K}(\a),\n_{L/K}(\a)\in\O_K$ if and only if $\a\in\O_L$.  One direction is clear.  For the reverse, observe
    that $\a$ is a root of the monic polynomial
    $$x^2-\tr_{L/K}(\a)x+\n_{L/K}(\a),$$
    and is therefore integral over $\O_K$ (and hence over $\Z$).  Since $\a\in L,$ we have $\a\in\O_L$.



    %On the other hand, if $\a\in\O_L,$ then
    %$\a$ satisfies monic polynomial with coefficients in $\Z$.


    %If we had $\a\in \O_L,$ then we would have
    %$$\disc(1,\a,\a^2)\in\Z,$$
    %since this latter is integral (being a determinant of integral things) and in $\Q$ (since it is fixed by any embedding
    %of $L$ into $\C$).  Using 9) we find that
    %\begin{align*}
    %\disc(1,\a,\a^2)&=\pm \n_{L/K}\left(3\a^2+\frac{1}{3}\right).\\
    %\end{align*}
    %Let the other roots of $f(x)$ be $\b,\c$.  Then we have
    %\begin{align*}
    %   \a+\b+\c&=0\\
    %    \a\b+\a\c+\b\c&=\frac{1}{3}\\
    %    \a\b\c&=1,
    %\end{align*}
    %and
    %\begin{align*}
    %    \n_{L/K}\left(3\a^2+\frac{1}{3}\right)&=\n_{L/K}\left(\frac{2\a^3+1}{\a}\right)\\
    %    &=\left(\frac{(2\a^3+1)(2\b^3+1)(2\c^3+1)}{\a\b\c}\right)\\
    %    &=8(\a\b\c)^3+2(\a^3+\b^3+\c^3)+4(\a^3\b^3+\a^3\c^3+\b^3\c^3)+1\\
    %    &=9+2((\a+\b+\c)^3-3(\a^2\b+\b^2\a+\a^2\c+\c^2\a+\b^2\c+\c^2\b)-6\a\b\c)+4((ac+ab+bc)^3-3())
    %\end{align*}

    \item Let $d$ be square free and $p$ a prime not dividing $2d$.  Set $K=\Q(\sqrt{d}).$  Since
    $\Delta_K=d$ or $4d$, we see that $p$ does not ramify in $K$.  Now $d$ is not a square modulo $p$
    if and only if $x^2-d$ has no solutions in $\Z/(p)$.  But in any case, $x^2-d$ has a root in $\O_K$
    (obviously) and hence has a root in $\O_K/\wp$ for any prime $\wp$ over $p$.  Therefore, $d$ is not a square modulo $p$
    if and only if
    $$\O_K/\wp \nsimeq \Z/(p).$$  Since $\deg(K/\Q)=2,$ it follows that
    $\O_K/\wp$ is a degree two extension of $\Z/(p)$ and hence that $d$ is not a square modulo $p$ if and only if $p$
    remains prime in $K$.

    \item Let $D$ be a dedekind domain with a finite number of prime ideals.  We show that $D$ is a PID.  Let
    $\p_1,\p_1,\ldots,\p_n$ be the distinct prime ideals of $D$.  Pick $\a_i\in p_i\setminus p_i^2$ such that
    $\a\not\in \p_j$ for $j\neq i$.  We know that such a choice is always possible by, for example, the chinese remainder
    theorem (problem 4).  Let us look at the prime decomposition of $(\a_i).$  Since $a_i\not\in\p_j$ for $i\neq j,$ we see that
    $\p_j\not|(\a_i).$  Similarly, since $\a_i\not\in\p_i^2,$ we have $\p_i^2\not| (\a_i)$.  It follows that $(\a_i)=p_i,$
    that is, each prime ideal is principal.  Hence, since $D$ is a dedekind domain and we therefore have unique factorization into prime ideals,
    every ideal of $D$ is principal.

    \item Define
    $$f:\O\longrightarrow \bigoplus_{i=1}^n \O/\ga_i$$ by
    $$f(x)=(x\mod \ga_1,x\mod \ga_2,\ldots,x\mod \ga_n).$$
    We claim that $f$ is surjective.  It is sufficient to show that we can obtain the element $(0,\ldots,0,1,0,\ldots,0)$ where
    the $1$ occurs in the $i^{\text{th}}$ place, for any $i$.  Since $\ga_i+\ga_j=\O$ for $i\neq j$, we have
    $$\ga_i+\bigcap_{j\neq i}\ga_j=\O.$$  To see this, we proceed by induction: let $\a\in \ga_i,\b\in\ga_j$ satisfy
    $\a+\b=1$.  Let $I\subset \O$ be any ideal such that $I+\ga_i=\O$ and $I+\ga_j=\O$, and let $\c\in I,\a^{'}\in \ga_i$
    satisfy $\c+\a^{'}=1.$  Then we have
    $$\a\c+\b\c=\c$$ and hence that
    $$\a^{'}+\a\c+\b\c=1.$$
    It follows that
    $$\a+\a^{'}\b+\a\c\b+\b^2\c=1.$$
    Since $\a+\a^{'}\b\in \ga_i$ and $\a\c\b+\b^2\c\in I\cap\ga_j$, we are done.  Therefore, there exists $\c\in \bigcap_{j\neq i}\ga_j$
    and $\a\in \ga_i$ such that $\a+\c=1$.  Then $\c\in\O$ satisfies $\c\mod \ga_j=0$ for $j\neq i$ and $\c\mod\ga_i=1$.  It follows
    that $f$ is surjective.  The kernel of $f$ is the ideal of all $x\in\O$ with $x\in \ga_i$ for all $i$.  That is,
    $$\ker f=\bigcap_{i=1}^n \ga_i:=\ga.$$
    We then have the isomorphism
    $$\O/\ga\simeq\bigoplus_{i=1}^n \O/\ga_i.$$

    \item Let $K=\Q(\sqrt{-5})$ and $L=\Q(\sqrt{-6})$.
    \begin{enumerate}
        \item We first show that $\O_K$ is not a UFD.  Consider the two factorizations
        $$2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})=6.$$  We claim that $2,3,(1\pm\sqrt{-5})$ are all irreducible
        elements of $\O_K$.  Suppose not.  Then, for example, we could write $2=(a+b\sqrt{-5})(c+d\sqrt{-5})$
        with $a,b,c,d\in \Z$ since $\O_K=\Z[\sqrt{-5}].$  Taking norms gives
        $$4=(a^2+5b^2)(c^2+5d^2),$$ from which it follows (since we are assuming our factorization
        of $2$ to be non trivial) that $a^2+5b^2=c^2+5d^2=2$.  But $2$ is not a square modulo $5$ and we have reached a
        contradiction.  Similar considerations show that if $3$ is not irreducible, then $3=a^2+5b^2$ for some $a,b\in\Z$,
        again a contradiction.  Now if $\O_K$ were a UFD then we would have $2|(1\pm\sqrt{-5})$.  In either case, this requires
        that $4|6,$ which is a contradiction.

        \item We use the Minkowski bound to determine the class numbers of $K,L$.  We have
        \begin{align*}
            \Delta_K&=-20,\\
            \Delta_L&=-24,
        \end{align*}
        and
        $$d_K=d_L=2.$$  Observe that $s=1$ in both cases.
        The Minkowski bound is then
        $$\frac{2!}{2^2}\left(\frac{4}{\pi}\right)\sqrt{20}<3$$
        for $K$ and
        $$\frac{2!}{2^2}\left(\frac{4}{\pi}\right)\sqrt{24}<4$$
        for $L$.  Therefore, every ideal class of $K$ contains an $\O_K$ ideal of norm at most $2$.  Therefore, we only need to
        consider those ideals of $\O_K$ of norm exactly $2$.  But any ideal of norm $2$ is a prime above $2$, and we know that
        $2$ totally ramifies in $K$ since it divides the discriminant.  Moreover, since $x^2-5\equiv (x+1)^2\mod 2,$
        we have (since $\O_K=\Z[\sqrt{5}]$),
        $$(2)=(2,1+\sqrt{5})^2.$$  Moreover, it is not difficult to see that $(2,1+\sqrt{-5})$ is not principal: if it were,
        we could write $2=a^2+5b^2$ for $a,b\in\Z$, which is absurd as we have already seen.  Thus, $\cl_{\O_K}$ consists
        of two elements and is hence isomorphic to $\Z/(2).$

        For $L$, we proceed as above and find
        \begin{align*}
            (2)&=(2,\sqrt{-6})^2\\
            (3)&=(3,\sqrt{-6})^2.
        \end{align*}
        As before, it is not hard to see that $(2,\sqrt{-6}),(3,\sqrt{-6})$ are not principal.  However, we have
        $$(2,\sqrt{-6})(3,\sqrt{-6})=(6,2\sqrt{-6},6\sqrt{-6},-6)= (\sqrt{-6}),$$
        so that if $[\alpha]$ denotes the ideal class of $\alpha\subset\O_L$, we have
        $$[(2,\sqrt{-6})]=[(2,\sqrt{-6})]^{-1},$$
        from which it follows that $\cl_{\O_L}\simeq \Z/(2).$
    \end{enumerate}

    \item Suppose that $K$ is a norm Euclidean imaginary quadratic field.  Then given any $\a,\b\in\O_K,$ there exists
    $q,r\in \O_K$ such that
    $$\a=\b q+r$$ and $$\n_{K/\Q}(r)<\n_{K/\Q}(\b).$$  Observe that $\O_K$ is a lattice in $\C$ and that the norm of
    any $\a\in K$ is just the square of its usual euclidean distance from the origin ($\a=u+v\sqrt{-D}=u+iv\sqrt{D}$ with $u,v\in \Q$,
    and $\n_{K/\Q}(\a)=u^2+dv^2$ while the distance of $(u,v\sqrt{D})$ from the origin is $u^2+Dv^2$).  Thus, $K$ is norm
    euclidean if and only if given $\a,\b$ as above we can find $q\in\O_K,t\in K$ such that
    $$\frac{\a}{\b}=q+t$$ where $\n_{K/\Q}(t)<1$.  Therefore, $K$
    is norm euclidean if and only if given any point $x$ of $\C$ (since any point of $\C$ can be approximated as closely as we like
    by an element $\a/\b$ of $K$ with $\a,\b\in\O_K$ as above) we can find a lattice point $q\in\O_K$ of distance less than $1$
    away from $x$.  Equivalently, $K$ is norm euclidean if and only if the minimum distance from any interior point of a fundamental
    domain to a lattice point on the boundary of that fundamental domain is strictly less than $1$.  We then have two cases.
    \begin{enumerate}
        \item $-D\not\equiv 1\mod 4:$  In this case, $\O_K=\Z\oplus\Z\sqrt{-D}.$  A fundamental cell $F$ for the lattice is the rectangle
        of side lengths $1,\sqrt{D}$.  Given a point $p=x+iy\in F$, let $d(p)=\min_{q\in\O_K}|p-q|^2$, where $|*|$
        denotes the usual complex absolute value.  Then it is obvious that
        $$\sup_{p\in F}d(p)$$ occurs when $p=1/2+\sqrt{-D}/2$.  In this case, $p$ is equidistant from the four neighboring lattice points and the
        distance to any of them is
        $$\frac{\sqrt{1+D}}{2}.$$
        This is less than $1$ if and only if $D=1$ or 2.

        \item $-D\equiv 1\mod 4:$ Here, $\O_K=\Z\oplus\Z\left(\frac{1+\sqrt{-D}}{2}\right).$  A fundamental cell $F$ for the lattice is the
        parallelogram with vertices
        $$\left\{(0,0),(1,0),(1/2,\sqrt{D}/2),(3/2,\sqrt{D}/2)\right\}.$$
        Let $d(p)$ be as above.  Again, it is not difficult to see that
        $$\sup_{p\in F}d(p)$$ occurs when $p=3/4+\sqrt{D}/4$.  In this case, we have
        $$d(P)=\frac{1+D}{16}.$$  It follows that
        we must have $D<15$.  Since we are requiring that $-D\equiv 1(4)$ the only possible choices are
        $$D=3,7,11.$$
    \end{enumerate}

    Thus, the only norm euclidean imaginary quadratic fields are
    $$\Q(\sqrt{-1}),\Q(\sqrt{-2}),\Q(\sqrt{-3}),\Q(\sqrt{-7}),\Q(\sqrt{-11}).$$
    It is not true that every quadratic imaginary field with class number one is norm euclidean.  For example, $\Q(\sqrt{-19})$
    has class number one but is not norm euclidean as the following counterexample shows:
    $$\a=3+\sqrt{-19},\b=4.$$  In this case, the closest lattice points to $\a/\b$ are $1$ and $\frac{1+\sqrt{-19}}{2}$.  But in one case we have
    $$\n_{K/\Q}(\a-\b)=20\nless \n_{K/\Q}(\b)=16$$ while in the other,
        $$\n_{K/\Q}\left(\a-\b\left(\frac{1+\sqrt{-19}}{2}\right)\right)=\n_{K/\Q}(1-\sqrt{-D})=20\nless \n_{K/\Q}(\b)=16.$$


    \item Let $\O$ be a dedekind domain and $\ga$ a nonzero ideal of $\O$.  Suppose that
    $$\ga=\p_1^{a_1}\ldots \p_r^{a_r}.$$  Then by the chinese remainder theorem we have
    $$\O/\ga\simeq \bigoplus_{i=1}^r \O/\p_i^{a_i}.$$  Suppose that every ideal of $\O/\p_i^{\a_i}$
    is principal and let $I\subset \O/\ga$ be an ideal of $\O/\ga$.  Then $I$ corresponds to an ideal $\tilde{I}\subset \O$
    containing $\ga$.  Clearly, $\tilde{I}/\p_i^{a_i}$ is an ideal of $\O/\p_i^{a_i}$ and is principal by assumption, say
    $$\O/\p_i^{a_i}=x_i\O/\p_i^{a_i}.$$  Then we see that $I$ is a principal ideal of $\O/\ga$ generated by
    $$(x_1,x_2,\ldots,x_r)\in \bigoplus_{i=1}^r \O/\p_i^{a_i}\simeq \O/\ga.$$  It therefore suffices to show that
    for any prime ideal $\p,$ every ideal of $\O/\p^m$ is principal.  Let $I$ be any ideal of $\O/\p^m$.  Then $I$ corresponds
    to an ideal $\tilde{I}$ of $\O$ containing $\p^m$.  But since $\p^m\subset \tilde{I}$  we have $\tilde{I}|\p^m$ and hence
    that $\tilde{I}=\p^n$ for some $n\leq m$.  If $n=m$ then $I$ is the zero ideal and obviously principal.  Otherwise we may assume that
    $n<m$.  Let $\a\in \p^n\setminus \p^{n+1}.$  Then we have the containments
    $$\p^n\supseteq \p^m+(\a)\supsetneq \p^m,$$
    since $\a\not\in \p^{n+1}\subseteq \p^m.$  We therefore must have
    $$\p^n|\p^m+(\a)|\p^m,$$ so that
    $$\p^m+(\a)=\p^i$$ for some $n\leq i<m$.  Suppose $n<i$.  Then $\a\in \p^m+(\a)=\p^i$.  But $\a\not\in \p^{n+1}$ and hence $\a\not\in\p^i$.
    It follows that $n=i$ and
    $$\p^n=\p^m+(\a).$$  Therefore,
    $$I=\a\O/\p^m$$ is principal.  We now show that every ideal of $\O$ is generated by two elements.  Let $I\subset \O$ be an ideal, and
    let $\a\in\O$.  We have seen that every ideal of $\O/(\a)$ is principal.  Since $(\a)\subset I,$ it follows that $I/(\a)$ is an ideal
    of $\O/(\a)$ and is hence principal, say $I/(\a)=\bar{\b}\O/(\a).$  We thus have $I=(\a,\b),$ where $\b$ is any lift of $\bar{b}$.

    \item Let $K$ be a number field, $p$ a prime number and $\p$ a non zero prime ideal of $\O_K$.  Set $E_r=\Q(\zr)$ and
    $L_r=K(\zr)$.  We will show that there exists a constant $c(\p)$ depending only on $\p$ (once $p$ is fixed) such that
    $\p$ splits into at most $c(\p)$ prime factors in $L_r$ for all $r$.  Consider the following diagram of field extensions:

    \setlength{\unitlength}{1in}
    \begin{center}
    \begin{picture}(3,3)
    \put(1.5,2.5){$L_r$}
    \put(2.5,1.5){$K$}
    \put(0.5,1.5){$E_r$}
    \put(1.53,0.5){$\Q$}

    \put(1.58,2.45){\line(1,-1){0.9}}
    \put(1.58,2.45){\line(-1,-1){0.9}}
    \put(0.68,1.54){\line(1,-1){0.9}}
    \put(2.48,1.54){\line(-1,-1){0.9}}
    \end{picture}
    \end{center}

    Since $L_r=E_rK,$ it follows that $\deg(L_r/E_r)\leq \deg(K/\Q)$.  Let $q=\p\cap\Z$.  We claim that it
    suffices to show that there exists a constant $c(q)$ such that $q$ splits into at most $c(q)$ primes in $E_r$
    for all $r$.  Indeed, let $\q$ be any prime of $E_r$ and write $\q=\wp_1^{e_1}\cdots\wp_s^{e_s}$ with $\wp_i$
    primes of $\O_{L_r}$.  Then we have $e_1f_1+\cdots+ e_sf_s=\deg(L_r/E_r)\leq \deg (K/\Q)$.  In particular,
    since $e_i,f_i\geq 1,$ we have $s\leq \deg(K/\Q),$ so that there exists some constant $C$ independent of $r$
    such that every prime of $E_r$ splits into at most $C$ factors in $L_r$.  We now show that there exists some
    constant $c(q)$ such that $q$ splits into at most $c(q)$ prime factors in $E_r$ for all $r>0$.  We distinguish two cases:
    \begin{enumerate}
        \item $p=q$.  Since $p$ totally ramifies in $E_r$ for all $r$, there is exactly one prime above $p$.  In this case,
        we see that $\p$ splits into at most $\deg(K/Q)$ primes in $L_r$ for any $r$.

        \item $p\neq q$.  Let $\q$ be a prime over $q$ in $E_r$.  Recall from lecture that $E_r/\Q$ is Galois and that
        $$e(\q/q)=1$$ and
        $$f(\q/q)=f_r$$ is the order of $q$ modulo $p^r$.  Since $e_rf_rd_r=\varphi(p^r)$, where $d_r$ is the number of primes
        lying over $q$ in $E_r$, we see that
        $$d_r=\frac{p^{r-1}(p-1)}{f_r}$$ for any $r$.
        Now let $r_0$ be such that
        $$q^{f_1}=1+cp^{r_0},$$
        where $p\not| c$.  Then $f_1=f_r$ for $1\leq r\leq r_0$.  Observe that
        $$q^{f_1p^{j-1}}\equiv 1+cp^{r_0+j-1}\mod p^{r_0+j}\not\equiv 1\mod p^{r_0+j}$$ but
        $$q^{f_1p^{j}}\equiv 1+cp^{r_0+j}\mod p^{r_0+j}\equiv 1\mod p^{r_0+j}$$ for any $j\geq 1$.
        It follows that $f_{r_0+j}|f_1p^j$ but $f_{r_0+j}\not|f_1p^{j-1}$, and hence (since $f_1|f_r$ for any $r\geq 1$) that
        $$f_{r_0+j}=f_1p^j.$$
        Therefore,
        $$d_{r_0+j}=\frac{p^{r_0+j-1}(p-1)}{f_{1}p^j}=\frac{p^{r_0-1}(p-1)}{f_{1}}$$
        for all $j\geq 1$.  This proves our claim.  Putting this analysis together, it is not difficult
        to see that
        $$c(\p)=\frac{p^{r_0-1}(p-1)}{f_1}\deg(K/Q),$$
        where $f_1$ is the order of $l=\p\cap\Z$ modulo $p$ and $r_0$ is such that
        $$l^{f_1}\equiv 1\mod p^{r_0}$$ but
        $$l^{f_1}\not\equiv 1\mod p^{r_0+1}.$$
    \end{enumerate}



    \item Let $L=K[\a]$ be a finite, separable extension and let $f$ be the minimal monic polynomial of $\a$ over $K$.
    Set $n=\deg(f)$ and let $\sigma_i$ for $i=1,\ldots,n$ be the $n$ embeddings of $L$ in $\C$.
    We have
    \begin{align*}
        \Delta(1,\a,\a^2,\ldots,\a^{n-1})&=\begin{vmatrix}1 & \a & \a^2 &\cdots & \a^{n-1}\\
        1 & \sigma_1(\a) & \sigma_1(\a)^2 &\cdots & \sigma_1(\a)^{n-1}\\
        \vdots &  &  &\cdots & \vdots \\1 & \sigma_n(\a) & \sigma_n(\a)^2 &\cdots & \sigma_n(\a)^{n-1}\end{vmatrix}^2\\
        &=\prod_{1\leq i <j \leq n}(\sigma_{i}(\a)-\sigma_j(\a))^2\\
        &=(-1)^{n(n-1)/2}\prod_{i\neq j}(\sigma_{i}(\a)-\sigma_j(\a)),
    \end{align*}
    where we have used the Vandermonde determinant identity in the second line.  Now we have
    $$f^{\prime}(\sigma_i(\alpha))=\prod_{j\neq i}(\sigma_i(\a)-\sigma_j(\a)),$$ so that
    $$\Delta(1,\a,\a^2,\ldots,\a^{n-1})=(-1)^{n(n-1)/2}\prod_{i=1}^n f^{\prime}(\sigma_i(\a))=(-1)^{n(n-1)/2}\n_{L/K}(f^{\prime}(\a)).$$

    \item Let $K_1=\Q(\a),K_2=\Q(\b),$ where $\a,\b$ are roots of $p=x^3-8x+15,q=x^3+10x+1$ respectively.  We show that $K_1,K_2$ have
    the same discriminant but that they are not isomorphic.  Let us first compute $\Delta(1,\a,\a^2)$ and $\Delta(1,\b,\b^2)$.
    It is clear that $p,q$ are irreducible since any rational root of $p$ would be an integer dividing 15, while a rational root
    of $q$ would be an integer dividing 1.  Neither $p$ nor $q$ have such roots.  Thus, by Exercise 9 we have
    \begin{align*}
        \Delta(1,\a,\a^2)&=-\n_{K_1/\Q}(3\a^2-8)\\
        &=-(3a_0^2-8)(3a_1^2-8)(3a_2^2-8)\\
        &=-(27a_0^2a_1^2a_2^2 - 72(a_0^2a_1^2 +a_0^2a_2^2+ a_1^2a_2^2)+ 192(a_0^2 + a_1^2 + a_2^2) - 512)\\
        &=-(27(a_0a_1a_2)^2-72((a_0a_1+a_1a_2+a_0a_2)^2-2(a_0+a_1+a_2)a_0a_1a_2)+192((a_0+a_1+a_2)^2\\
        &-2(a_0a_1+a_0a_2+a_1a_2)))\\
        &=-(27(-15)^2-72((-8)^2)+192(-2(-8))-512)\\
        &=-4027,
    \end{align*}
    where $\a=a_0,a_1,a_2$ are the roots of $p$.  Similarly, we compute
     \begin{align*}
        \Delta(1,\b,\b^2)&=-\n_{K_2/\Q}(3\b^2+10)\\
        &=-(3b_0^2+10)(3b_1^2+10)(3b_2^2+10)\\
        &=-(27b_2^2b_1^2b_0^2 + 90(b_0^2b_1^2 + b_0^2b_2^2+b_1^2b_2^2)+ 300(b_0^2 +b_1^2+b_2^2) + 1000)\\
        &=-(27(b_0b_1b_2)^2 + 90((b_0b_1+b_1b_2+b_0b_2)^2-2(b_0+b_1+b_2)b_0b_1b_2)+ 300((b_0+b_1+b_2)^2\\
        &-2(b_0b_1+b_0b_2+b_1b_2)) + 1000)\\
        &=-(27(1)+90(10^2)+300(-20)+1000)\\
        &=-4027.
    \end{align*}
    Now observe that $4027$ is prime so that, in particular, it is square free.  It follows that
    $$\O_{K_1}=\Z[\a]$$ and
    $$\O_{K_2}=\Z[\b],$$
    and hence that $K_1,K_2$ have the same discriminant (-4027).  However, $K_1,K_2$ are not isomorphic.  For
    consider the factorization of $17\O_{K_i}$ in $\O_{K_i}$ for $i=1,2$.  We have (as is easily verified)
    \begin{align*}
        x^3-8x+15&\equiv (x+4)(x+6)(x+7)\mod 17,
    \end{align*}
    while $x^3+10x+1$ is prime modulo 17.  It follows that $17$ is totally split in $\O_{K_1}$ but inert in
    $\O_{K_2}$; it follows that $K_1$ cannot be isomorphic to $K_2$.

    \item Let $K=\Q(\zr)$.  We have seen that $\O_K=\Z[\zr]$ so that
    $$\Delta_K=\Delta(1,\zr,\zr^2,\ldots,\zr^{\varphi(p^r)-1})=\n_{K/\Q}(f^{\prime}(\zr)),$$
    where $f$ is the minimal monic polynomial of $\zr$ over $\Q$.  We have also seen that
    $$f(x)=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}.$$
    We therefore have
    \begin{align*}
        p^r x^{p^r-1}&=(x^{p^{r-1}}-1)f^{\prime}(x)+(p^{r-1}x^{p^{r-1}-1})f(x),
    \end{align*}
    so that
    \begin{align*}
        f^{\prime}(\zr)=\frac{p^r \zr^{p^r-1}}{\zr^{p^{r-1}}-1}=\frac{p^r}{\zr(\zr^{p^{r-1}}-1)}.
    \end{align*}
    Therefore, we have
    $$\n_{K/\Q}(f^{\prime}(\zr))=\frac{\n_{K/Q}(p^r)}{\n_{K/Q}(\zr)\n_{K/Q}(\zr^{p^{r-1}}-1)}.$$
    Clearly, $\n_{K/\Q}(\zr)=1$ and
    $$\n_{K/Q}(\zr^{p^{r-1}}-1)=\n_{K/Q}(\z_p-1)=\left(\n_{\Q(\z_p)/\Q}(\z_p-1)\right)^{\deg(K/\Q(\z_p))}=p^{p^{r-1}},$$
    since
    $$\n_{\Q(\z_p)/\Q}(\z_p-1)=\prod_{(p-1,j)=1}(1-\z_p^j)=1+1^2+\cdots+1^{p-1}=1.$$
    Therefore,
    \begin{align*}
    \n_{K/\Q}(f^{\prime}(\zr))&=\frac{p^{r\varphi(p^r)}}{p^{p^{r-1}}}\\
    &=p^{rp^{r-1}(p-1)-p^{r-1}}\\
    &=p^{p^{r-1}(pr-r-1)}.
    \end{align*}
    It follows that
    $$\Delta_K=\pm p^{p^{r-1}(pr-r-1)}.$$

    \item Using Minkowski, we find that every $\O$ ideal is equivalent to some ideal $J$
        with $\n(J)\leq 3$.  In this case, $\O=\Z[\frac{1+\sqrt{-23}}{2}]:=\Z[\alpha]$, and we look
        at the irreducible polynomial
        \begin{eqnarray*}
            X^2-X+\frac{1+23}{4}=X^2-X+6
        \end{eqnarray*}
        modulo $2,3$.  This gives the factorizations
        \begin{eqnarray*}
            (2)&=&(2,\alpha)(2,\alpha-1)=P_2\bar{P_2}\\
            (3)&=&(3,\alpha)(3,\alpha-1)=P_3\bar{P_3}.
        \end{eqnarray*}
        Since $\frac{1+\sqrt{-23}}{2}\frac{1-\sqrt{-23}}{2}=6$ and the two factors are conjugate, we must have
        \begin{eqnarray*}
            \left(\frac{1+\sqrt{-23}}{2}\right)=P_2P_3
        \end{eqnarray*}
        up to relabeling.  Therefore, $\cl_{O}$ is generated by $[P_2]$.  Now, the factorization
        \begin{eqnarray*}
            \frac{3+\sqrt{-23}}{2}\frac{3-\sqrt{-23}}{2}=2^3
        \end{eqnarray*}
        tells us that
        \begin{eqnarray*}
            \left(\frac{3+\sqrt{-23}}{2}\right)=P_2^3
        \end{eqnarray*}
        up to relabeling and units (it cannot be that
        only a single or double power of $P_2$ divides $\frac{3+\sqrt{-23}}{2}$, for then we would have
        $\left(\frac{3+\sqrt{-23}}{2}\right)=P_2^2\bar{P_2}$, up to relabeling, which is impossible
        since $[P_2]^2[\bar{P_2}]=[P_2]$ is not principal).  Therefore, $[P_2]^3=1$ so that
        $[P_2]$ must have order $3$.  It follows that
        \begin{eqnarray*}
            \cl_{\O}\simeq \Z/3\Z
        \end{eqnarray*}
        and that the three ideal classes are represented by
        $$1,P_2,P_2^2.$$

    \item Let $K$ be a number field and $\m\subset \O_K$ an ideal.  We show that every ideal class of $\cl_{\O_K}$
    contains an integral ideal $\ga$ such that $(\m,\ga)=1$.  For any ideal class $\gc,$ let $J\in \gc$ be an integral ideal
    representing $\gc$ (recall that we showed in class that such a $J$ exists).  Let
    $$\wp_1,\ldots,\wp_n$$ be the primes (if any) dividing $\m$ that do not divide $J$.  Further, write
    $$J=\q_1^{e_1}\cdots\q_s^{e_s}.$$
    By the chinese remainder theorem, we can find $\alpha\in\O_K$ such that
    \begin{align*}
        \a &\in \q_i^{e_i}\setminus \q_i^{e_i+1}\\
        \intertext{and}
        \a &\equiv 1\mod \wp_j
    \end{align*}
    for all $i,j$.  Then we clearly have
    $$(\a)=\prod_{i=1}^s \q_i^{e_i}\prod_{j=s+1}^{s+t} \p_j^{e_j}:=J\ga$$
    with $\q_i\not|\ga$ for any $i$.  By construction, $\p_j\not|\m$ for any $j$
    since if $\p_j|\m$ for some $j$ then we have a prime not dividing $J$ that divides $\m$
    and $(\a)$, but we chose $\a$ to be 1 modulo all of these primes.
    Therefore, $\ga$ is an integral ideal in the class of $J^{-1}$ (since the product $J\ga$ is
    principal) with $(\m,\ga)=1.$  It follows that every ideal class contains an integral ideal
    $\ga$ with $(\m,\ga)=1$ since every class is the inverse of some other class.

\end{enumerate}

\end{document}
